How to prove $p(a \cap \bar b)=p(a)+p(a\cup b)$?
I tried substitution for $\bar b$, but that did not work.
I know of the identity $p(a \cap \bar b)= p(a)+p(a \cup b)$.
I suspect deMorgans laws is in there but not sure how to apply it.
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5What you want to prove is palpably false: both terms on the right are individually larger than the term on the left and hence so is their sum. – Dilip Sarwate Feb 24 '18 at 13:54
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What are you complaining about @Dilip Sarwate ? This gives a new way to break the unity barrier in probability. True, the unity barrier has been broken before, but it's always interesting to see new ways to do it https://stats.stackexchange.com/questions/4220/can-a-probability-distribution-value-exceeding-1-be-ok/160979#160979 . – Mark L. Stone Feb 24 '18 at 19:00
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I possibly wrote it wrong . It was probably p( a n b ‘)= p(a) - p(a n b) – larry mintz Feb 24 '18 at 22:04
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@larry mintz Use a Venn diagram. – Mark L. Stone Feb 24 '18 at 22:32
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The closest thing that I can think of to fix your equality is
$$p(a \cap \bar{b})=p(a)-p(a \cap b)$$
of which it can be easily obtained by partitioning $a$ into subsets which contain $b$ and doesn't.
But clearly $-p(a \cap b) \neq p(a \cup b)$.
Siong Thye Goh
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I wrote it down wrong. I know of the identity n(a n b’)=n(a)-n(a n b) – larry mintz Feb 24 '18 at 22:00
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1just use the property that $a \cap \bar{b}$ and $a \cap b$ are mutually exclusive and finite additivity. – Siong Thye Goh Feb 25 '18 at 04:41